Let $p:X\rightarrow Y$ be a fiber bundle with fiber $F$ that is contractible.
Clearly $\pi_n(X)\cong \pi_n(Y)$ for all $n$. But why does it follows that $H^{*}(X)\cong H^{*}(Y)$?
Is it because the iso $\pi_n(X)\cong \pi_n(Y)$ is induced by a map?
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Yes in the setting of CW complexes, see answers to duplicates to your previous question. Probably the same in general. – Moishe Kohan May 21 '23 at 19:25
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@MoisheKohan what about things that are not CW complexes? – May 21 '23 at 19:35
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Theorem 7.6.25 (page 402) in
Spanier, Edwin H., Algebraic topology. (1st corr. Springer ed. of the orig. publ. by McGraw- Hill, 1966), New York - Heidelberg - Berlin: Springer-Verlag. XIV, 528 p. (1982). ZBL0477.55001.
states that a weak homotopy equivalence $X\to Y$ of topological spaces induces isomorphisms of homology groups $H_\cdot(X)\to H_\cdot(Y)$ (alternatively, see Whitehead's Theorem 7.5.9, page 399). By the UCT we will obtain the same for cohomology groups.
Moishe Kohan
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Thanks so much. Could you please tell me the page number? My edition of the book ends at page 528. – May 21 '23 at 21:32
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@MathematicallyInterested: Done. I think we are using the same edition from 1981. – Moishe Kohan May 21 '23 at 21:38
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